point-set-topology

[empty grove]

Then this is the answer, not sure if I phrased it properly then

[kind cedar]

I have many small blocks of information but can't fit it all together. Top spaces, metric spaces, convergence, completeness, Hausdorff... it'll take me a while to completely understand and separate this concepts

[empty grove]

Right, will help to just look at a bunch of examples and non examples of each

[kind cedar]

Thank you mate. I'll come back later with more questions (smarter ones I hope) ahaha

[empty grove]

[kind cedar]

Ah, again on the integers.... the positive integer line is Hausdorff? Because if x=2 and y=3, I don't see how I can get 2 disjoint open sets containing them both

[empty grove]

Balls of radius half are open

And they are just singletons

Like if you just look at definition of ball of radius r

It doesn't say there needs to be a point at all possible distances less than r

[kind cedar]

I'm not using the standard topology

based on balls

I'm trying to think of open sets in the topology, without using the ball definition

But, yeah, then I didn't define the topology for the space I'm dealing with

[empty grove]

The metric topology is defined by balls 🏀

[kind cedar]

Forget the metric for a while

[empty grove]

ah

[kind cedar]

I'm back to just hausdorff and integers

trying to understand Hausdorff

[empty grove]

So what's the topology that you are taking on the integers?

Hausdorffness depends on the topology

[kind cedar]

Let's suppose the discrete topology to start with

[empty grove]

Then {x}, {y} form a separation of x and y by disjoint open sets

[kind cedar]

even if x and y differ by one?

[empty grove]

Yes, discrete topology means everything is open

[kind cedar]

well, if all sets are open, then {x} is open and so is {y} and they are disjoint. ok

then the definition that a single point in a hausdorff space being closed is contradictory

[empty grove]

I didn't get that

[kind cedar]

If we just understood the integers to compose a Hausdorff space

Since any 2 points have a disjoint open around them

[empty grove]

open doesn't mean not closed if that is the confusion

[kind cedar]

AH, of course!

Yeah, you're right

[empty grove]

[kind cedar]

So... in the discrete topology, all sets are also closed?

[empty grove]

Yep

[kind cedar]

Since they're all complements to open

Oh god

[empty grove]

Yes

[kind cedar]

Now I get why this topology is useless hahaha

ok, gotta get back here, thank you again

[marsh forge]

(its not useless for the record)

[kind cedar]

Please, elaborate further

[marsh forge]

There are many spaces for which the discrete topology is the most natural one

It is also an important formal thing in category theory of top

[gritty widget]

A lot of counter-examples of uniform spaces are just discrete spaces with neatly chosen uniformities

[ornate lark]

anyone familiar with lax pairs?

[gritty widget]

yooo monikernel long time no see

[gritty widget]

@gritty widget leave me alone

why did you leave

[hallow swan]

why is it that in the metric topology, if x_0 \in U where U is some open subset, then there exists some ball centered at x_0 contained in U

[gritty widget]

in metric topology open balls form basis

In particular if x_0 is in U then there's some open ball it belongs to

then find such ball centred at x_0, contained in U

[hallow swan]

oh i forgot the traditional definition of basis

i was thinking a set U is open if it is expressible as a countable union of basis elements

but the other (equivalent) definition makes this obvious

thanks

[gritty widget]

this is almost true. The union doesn't have to be countable

[hallow swan]

oh arbitrary unions?

[gritty widget]

For example taking R with discrete topology, sets of the form {x} form a basis and any uncountable set is uncountable union of basis elements

[hallow swan]

I see

[kind cedar]

Another stupid question: this \ {0} means "excluding zero" ?

[limpid leaf]

Yes

How do I show the second condition of this being a basis? For $f \in C([0,1]$ and $\varepsilon >0$, let $U(f, \varepsilon) = {g \in C([0, 1]) \colon \int_0^1 |f(x) - g(x)| < \varepsilon}$. Then the collection of $U(f, \varepsilon)$ is a basis for a topology on $C([0,1])$. I was sort of able to do something like
let $h \in U(f, \varepsilon_1) \cap U(g, \varepsilon_2) \Rightarrow \int_0^1 |[f(x) + g(x) - h(x)] - h(x)|$ and let $k = f(x) + g(x) - h(x)$, $h(x) \in U(k, \varepsilon_1 + \varepsilon_2)$, but I'm not sure if that's the right choice for $k$

[gentle osprey]

jesse

[limpid leaf]

well assuming I did what I did so far right. I think I've shown that there is a thing that contains stuff in teh intersection

is that all

i feel like I missed something

[gritty widget]

Oh, I'm pretty sure I did this exercise in Dugundji book

I was getting constantly stuck just like you

[little hemlock]

I’m guessing you haven’t proved there is a metric on C[0, 1]? Because the proof of this is the same in every metric space

[limpid leaf]

Yeah I figure this is a metric but no we haven't discussed metric spaces

[gritty widget]

sorry it was a different exercise but similar

[gritty widget]

well, since it's a metric space like kxrider said, you can just take k = h and epsilon small enough

[finite heath]

Hello - may I ask a question, or is someone else getting help rn :[]?

(no rush)

[limpid leaf]

yeah i guess i ma overcomplicating this

go ahead Toucan

[finite heath]

Yes sir

Thank you!

Context: New to proof based math | really struggling in topology course | will be in here constantly =[

Problem

So, I get that: for the plane to be open in R^2, there must be a ball centered at each point in the plane, where each ball is also entirely contained inside the plane

[unreal stratus]

Something I'd do is find the length of a given point from the line Ax + By = C

Do you see why that'd help geometrically?

[finite heath]

let me try and see sir

[gritty widget]

The easiest way for me to show this would be to show that the complement is closed

[finite heath]

oooo

[gritty widget]

and I'd do this using that a set in a metric space is closed if for any converging sequence from it, the limit is also in that set

[finite heath]

This is beyond my knowledge atm

I have never written a proof without assistance :/

I am very new

Hard for me to take your guy's advice and put it into work

[unreal stratus]

Well it's the distance from the line that matters actually

[finite heath]

But what I did above hopefully helps

[unreal stratus]

Which is a in your example

[finite heath]

the distance, a :o?

[unreal stratus]

Ye

So the idea is if you have an open ball radius, say, a

That'll be contained in the open half plane too

The case for general A B and C is basically the same

:)

[finite heath]

wdym?

I just set B and C to zero to make it visually simple

[unreal stratus]

yeah

I'm just saying the same proof works in general

[finite heath]

ooo okay

[unreal stratus]

By rotations etc

[long hornet]

You can also define f(x, y) = Ax + By - C and observe that your set is the preimage of (-inf, 0), which is open, so it is open

[finite heath]

Now, by 'the proof', i don't know how to do the proof

I have been stuck on this problem for hours

[unreal stratus]

Nice

[finite heath]

I don't know how to go about proofs really

this class is very challenging

[unreal stratus]

I'm sorta surprised you're doing topology if you're not used to proofs - are you studying another subject out of interest?

[finite heath]

I am majoring in applied math

ive done a ton of calculus

and some linear algebra

[unreal stratus]

Ah, fair yeah makes sense

[finite heath]

never any proofs

i wanted to see how a proof-based class would be

and boy

do i regret that

this aint nothing like what ive done

[unreal stratus]

Hopefully it'll be helpful and worth it tho?

Hm

[finite heath]

It will be worth it

totally blowing my mind

ive already learned a lot, but as i said, i need to be walked through every proof before i can get the hang of it

I would compensate you for more in-depth tutoring... I've tried so many resources

Chegg, Discords, Stack Exchange

The textbook

im desperate lol

down bad rn

[unreal stratus]

I'm definitely not the one who should be tutoring lol, nor do i have time unfortunately oop

Sorry

But yeah best of luck :)

[finite heath]

do u know any good resources :p?

Like

I just straight up dont know how to do this proof

How do I learn?

[unreal stratus]

It sounds like you'd like smth just for general proof writing/strategies?

I'd recommend looking at any uni's intro to proofs / intro to uni math(s) courses if possible i guess

[little hemlock]

@limpid leaf did you ever figure out your problem?

[limpid leaf]

no, i have something else to do but ill prob return to it in an hour or so

[little hemlock]

aight you can ping me if you want a hint or something

[pastel linden]

literally just started reading guillemin and pollack

why does he not mention smoothness of transition maps in the definition of a smooth manifold?

[gritty widget]

what's their definition of a smooth manifold

[pastel linden]

subset of R^n such that every point has a coordinate neighborhood in the subspace topology

To my understanding this is also assuming that all smooth manifolds embed in euclidean space

Take me back to Tu

Oh and a smooth function on a set is defined as a function that can be extended to a smooth function on an open set containing the set in the embedding space R^n, in which case smoothness js defined in terms of partials

[gritty widget]

smoothness of the transition maps might follow from their definition

i remember something similar in spivak's com

[pastel linden]

cant wait to relearn basic manifold stuff with slightly different defs

I will think about this

[fair idol]

Is there a classic example of a function which is only finitely differentiable?

[gritty widget]

something like x^2 sin(1/x) for x != 0 and 0 for x = 0?

if you wanted to find a function with k derivatives but not k + 1 derivatives, you could keep taking antiderivatives of something like the absolute value function

[fair idol]

Ah I see. That weird oscillating boy, of course

[orchid forge]

related: there are smooth (infinitely differentiable) functions which are not analytic (locally equal to a Taylor series) anywhere https://www.chebfun.org/examples/stats/Smoothies.html

(for a less extreme but simpler example, every derivative of e^(-1/x) approaches 0 as x --> 0, so extending it by 0 for x <= 0 gives a smooth function whose Taylor series at x=0 is 0)

[orchid forge]

This holds in the usual definition by the Whitney embedding theorem, although not necessarily in an obvious way (an n-dimensional manifold might embed into R^n or R^(2n), or anywhere in between)

the fact that the Whitney embedding is smooth is required to transfer the smooth structure of R^(2n) to the n-dimensional manifold

by definition, an embedding M --> R^(2m) is an isomorphism from M onto its image in the subspace topology, so it makes sense to say "each point has a coordinate neighborhood in the subspace topology"

finally, an embedding induces an injective linear map on the tangent space at each point, so the coordinate neighborhood at a point pulls back to a coordinate neighborhood in M

if read carefully, each of these is an equivalence, so that definition is equivalent to the standard one

[shy moss]

is every n-simplex simply connected?

[coral pawn]

Let X be a topological space with an equivalence relation ~. Let Y be a subspace of X and let ~^* be the induced equiv relation on Y. Prove that Y/~^* is homeomorphic to Y/~ where Y/~ is given the subspace topology

[shy moss]

i think that it is

[coral pawn]

Using universal properties

[icy schooner]

https://math.stackexchange.com/questions/2271218/example-of-connected-but-not-path-connected-set?rq=1

if anyone can help me understand this part
"As LN is connected (it is homeomorphic with [0,1]), we conclude that Ln⊆U."

how do we conclude Ln is in U?

[orchid forge]

more than that, it's homeomorphic to an n-dimensional ball

[honest terrace]

Ln is connected, and we know Ln \cap U \cap V = \emptyset and Ln C U \cup V

So either Ln C U or Ln C V, because Ln is connected

Now since a point of Ln is in U

And that Ln \cap U \cap V = \emptyset

That point can't be in V

So Ln must be included in U

Is this clearer ?

[icy schooner]

Yep thank you

[fair idol]

Manifolds are of course locally eyclidean by definition. Does this local homeomorphism property imply the existence of a chart on the manifold?

I'm a dunce I originally wrote locally homeomorphic instead of locally euclidean

Rats I said that wrong and have tried to fix what I wrong.

I mean manifolds are locally euclidean. Does this locally euclidean property imply the existence of charts on a manifold?

Sorry if this is confusing

Exactly what you said previously

It seems like it is exactly the same but I want to be sure

Okay thank you

[limpid leaf]

i would like this hint now xd

[little hemlock]

Define $d(f,g) = \int_0^1 |f(x)-g(x)| , dx$ for any $f,g \in C[0,1]$. If $h \in U(f, \epsilon_1) \cap U(g, \epsilon_2) $, we gotta show there is $\delta > 0$ such that if $d(h,k) < \delta$ then $d(k, f) < \epsilon_1$ and $d(k,g) < \epsilon_2$, right

[gentle osprey]

kxrider

[limpid leaf]

yes

yes okay i see what you are saying

wait dont tell em anymore this is good hint

thank u

[little hemlock]

haven't said much of anything yet but np ig

[limpid leaf]

i dont think i understood the setup

or it was jumbled in my mind

[little hemlock]

ah okay

[fair idol]

Given a chart (U,f) of some point on a manifold M on dimension k , is U uncountable?

I know an open set in R^k is uncountable and feel as though the homeomorphism transfers this but can't quite say.

[pastel linden]

if U is in bijection with an open subset of euclidean space, is U uncountable?

homeomorphisms are in fact bijections

[limpid leaf]

okay maybe i need another push

I cant figure out the choice of k

I thought it would just be like h(x) + min(e1,e2)/2? or something

[little hemlock]

oh we aren't choosing k

only thing we get to choose is delta. shoulda made taht clear mb

[limpid leaf]

oh okay

[little hemlock]

the idea is that we want U(h, delta) \subset U(f, epsilon_1)\cap U(g, epsilon_2).
I was just unpacking what this means

[limpid leaf]

yeah

[gritty widget]

when k > 0 :^)

[little hemlock]

you're probably familiar with the triangle inequality on the reals, right? Think about how you could apply it to the integral of the absolute value of the difference of two functions. That'll help you get upper bounds for d(k,f) and d(k,g).

[limpid leaf]

right yeah I figured it would be something like int(|f - k|) = int(|f - h + h - k|) \leq int(|f - h|) + int(|h - k|)?

Why do you write it like d(k, f)? I know it's symmetric but it seems you are meaning something else

[little hemlock]

because i don't want to write out the integral every time

[limpid leaf]

No, I mean instead of d(f, k)

we are showing k \in U(f, e_1), no?

[little hemlock]

oh, okay, no particular reason.
and yes

[limpid leaf]

okay i figured as much

[limpid leaf]

if I do this i get int(|f - k|) < e_1 + d? I am doing something wrong maybe

[little hemlock]

yep, thats right

[limpid leaf]

But that doesn't mean k \in U(f, e_1) I thought?

unless delta is <= 0

maybe for this case you could get a delta that works but idk how it would also work for int(|g - k|)

[little hemlock]

well okay so the idea is really that
d(f,k) <= d(f,h) + d

if you bound by epsilon youve gone too far

[limpid leaf]

like if you had d = min(e_1, e_2) - d(f, h) then you could have e_1 + min(e_1, e_2) - int(|f - h|) < e_1 + min(e_1, e_2) - e_1 \leq e_1?

oh

[little hemlock]

note we can define delta in terms of d(f,h) since d(f,h) is independent of our choice of k

[limpid leaf]

so is what I said on the right track? or

[little hemlock]

ur close but what you have written d = min(e_1, e_2) - d(f, h) doesn't quite work. I don't think we can guarantee this is > 0

[limpid leaf]

oh maybe like d(f, h)/2 would work?

[little hemlock]

well that would give d(f,k) <= 3e_1/2 at best

[limpid leaf]

yes

i cant think of delta that guarantees it wont be zero and works

[little hemlock]

we want d(f,k) <= d(f,h) + d < e_1 and d(g,k) <= d(g,h) + d < e_2

[limpid leaf]

yes

[little hemlock]

|| solve for d||

[limpid leaf]

of course

lmao

i didntthink of that

[little hemlock]

this works because d(f,h) < e_1 so e_1 - d(f, h) > 0

[limpid leaf]

right

[little hemlock]

so whats our choice of delta?

[limpid leaf]

[d(f, h) + d(f, g) - (e_1 + e_2)]/2?

no shot

[little hemlock]

we just want both of these inequalities
d(f,k) <= d(f,h) + d_1 < e_1 and d(g,k) <= d(g,h) + d_2 < e_2
to work out. The standard trick is to take d = min(d_1, d_2).

[limpid leaf]

oh so you're saying d = min(e_1 - d(f,h), e_2 - d(g, h))?

right yes that works of course

[little hemlock]

nice

[limpid leaf]

okay and so then d(f, k) < e_1 and d(g, k) < e_2 and so k \in d(f, k) cap d(g, k) and we are done?

Okay I just want to make sure of one thing which is that we are assuming d(h, k) < d right?

or does this follow from the definition of h and delta somehow

[little hemlock]

d(h,k) < d iff k is in U(h, d)

[limpid leaf]

right sorry my confusion was with the fact that we have U(h, d) being an element of the basis but I now see this is clear

okay awesome

thank u so much

you rock

[little hemlock]

ye npnp.

[gritty widget]

Let M be a manifold and G a lie group acting smoothly on M

Let V be a vector bundle on M

is there a "canonical" way to lift the action on M to an action on V

so we could define g*(x,v) by (gx,v)

or we could define g*(x,v) by (gx, gv) and chose some representation

[modest agate]

how can I go on to show that if $(X,D)$ is an anti-metric space (same as a metric space, the only difference is that it has the opposite triangle inequality) where $X$ is a set and $D:X\times X \to \mathbb{R}$ is a function then $X$ cannot have prime number of elements?

[gentle osprey]

notsushY

[pearl holly]

okay well you have D(x, x) = 0 right? Now try to use the opposite triangle inequality

[viral atlas]

If you see cl(A) as the set of all accumulation points of A, would you not expect them to be in cl(A\cup B)?

[modest agate]

yes, but how does it show that our X cannot contain prime elements?

[pearl holly]

the result is even stronger than that

oh wait

you mean "can not have prime cardinality"?

[modest agate]

I think so, it just said prove that X cannot have a prime number of elements

[pearl holly]

okay then take this as a hint

[fair idol]

On a topological manifold with a chart (U,f) is U necessarily uncountable by virtue of being homeomorphic to an open euclidean set?

[elder yew]

homeomorphisms are bijections, so yes

[pearl holly]

So Hatcher wants to prove this and h is just cohomology

This is the argument:

[empty grove]

I haven't read a single page since last meeting

[pearl holly]

I don't really understand this. It feels like Hatcher claims that $C^n(\coprod_\alpha X_\alpha) = C^n(\oplus_\alpha X_\alpha)$ and then uses the fact that $C^n(\oplus_\alpha X_\alpha) = \prod_\alpha C^n(X_\alpha)$. But how then is $\oplus$ defined on spaces? I must be missing something

[gentle osprey]

Tokidoki ✓

[pearl holly]

lmao it's okay, you have a lot of other stuff to do

[empty grove]

I think A_alpha are groups not spaces

[pearl holly]

ye right but what is the argument then?

[empty grove]

Universal property of direct sums

So maps out of direct sums are the same as a collection of maps, one from each factor

Same in the sense that there's a unique correspondence again

You can try proving this it shouldn't be too hard This says that the direct sum is the coproduct in the category of ab groups

[pearl holly]

yeah okay I know that but I don't really understand how Hatcher uses this to prove (3)

[empty grove]

oh to prove 3

[pearl holly]

So like when I read "which implies that the cochain complex of a disjoint union..."

[pearl holly]

I think of this

and here X_\alpha are spaces

[empty grove]

oh

Ye he isn't doing ⊕ X_a

He's taking chain complexes of each X_a and then taking direct sum

And disjoint union takes direct sum of your chain complexes

Because you can just write a chain in the union as a finite sum of chains in each X_a

[pearl holly]

wait what do you mean by this?

[empty grove]

Whoops I can't read

But he says

Direct product of the cochain complexes of the individual X_a

So he's not doing ∏ X_a

He's taking cochain complexes first

Then everything is abelian group so it makes sense to take direct product

[pearl holly]

okay wait let me think lmao

I still don't really see where the ∏ X_a comes in tho

maybe we can just take this on our next meeting where we can draw some stuff?

wait is $\text{Hom}(C_n( \coprod_\alpha X_\alpha), G) = \text{Hom}(\oplus_\alpha C_n(X_\alpha), G)$?

[gentle osprey]

Tokidoki ✓

[pearl holly]

man that is so ugly lmao

[empty grove]

I'm saying it doesn't

There is no product of spaces at any stage

[pearl holly]

wtf so how do you prove it then?

[empty grove]

Yes

[pearl holly]

oh okay then I see lmao

[empty grove]

Because those things are isomorphic before you even apply the hom functor

[pearl holly]

yeee right okay now I see. Thank you so much!

[empty grove]

[fair idol]

Are nontrivial open sets in Rn necessarily uncountable?

[pearl holly]

open balls do be uncountable tho 😌

[fair idol]

Trueeeeee.

[shy moss]

in the case X=S^1 v S^1 i think i can descompose T_f in the two S^1xI's under relation

then the intersection of the two S^1xI under relation is just the preimage of the basepoint right?

preimage under f

and i think that the fundamental group of each S^1xI under relation is Z^2 because it deformation retract onto torus, is this right?

[gritty widget]

Yeah in fact they are of size continuum. A lot of descriptive set theory is about proving "continuum hypothesis" for special subsets of, in particular, R^n. That is, they're countable or of size continuum.

For example closed and in general Borel sets of R^n

[novel acorn]

Math made me burnt out and I switched over to reading some physics so I never showed to any meeting even thought I told Toki I would 😔

Still am on 2.2 lmao
I got a little demotivated when I couldn't really do any problems well

[empty grove]

oof, happens

ask toki for lectures later

[novel acorn]

I could 👀
Tho I think I should practice my proofing more?
I dunno I'm just shit at these problems

[empty grove]

yeah solving exercises always good

[novel acorn]

Yeah
I mean they're fun when you actually solve them lmao

Also maybe I should start taking notes while reading

[marsh forge]

You won’t get the same answer for every map f

So no this is not correct

I didn’t understand your argument either though

[marsh forge]

This sounds wrong

But I didn’t quite catch your meaning

[fair idol]

Thank you! This is helpful

[shy moss]

T_f is the quotient space (S^1 v S^1) x I/~ and S^1 v S^1 is the disjoint union of two circles identifying a point, then we can descompose T_f in two S^1 xI/~

[marsh forge]

Okay sure

But you cannot work with them independently

[shy moss]

why?

[marsh forge]

Consider the map that swaps the circles

[shy moss]

in this case T_f is like a mobius band?

[marsh forge]

No?

Oh

I mean

Yes

I see what you mean

[shy moss]

so, can i descompose $(S^1 \vee S^1) \times I/\sim$ in the two $S^1 \times I/\sim\cup (S^1 \vee S^1,0)$?

[gentle osprey]

Or x1

[marsh forge]

What will you do next

[shy moss]

apply van kampen

[marsh forge]

You can split S1vS1 x I into two cylinders but the mapping torus cannot necessarily be decomposed that way. For example, if we take the swap map from earlier, than the two cylinders are connected to eachother by ~

Indeed separating at the wedge point x I would give you one big torus not two separate things

[shy moss]

but if we attach (S^1vS^1,0) to the cilynder then there are two separete things right?

[marsh forge]

Can you use more words

[shy moss]

we can descompose the space $(S^1 \vee S^1) \times I/\sim$ in the two cilynders union the wedge which is in the base of the cilynder $S^1 \times I/\sim\cup \lbrace (a,0)| a \in S^1 \vee S^1 \rbrace$

[gentle osprey]

Or x1

[marsh forge]

Yes I think I get your meaning now

You can do this to apply van kampen yes, although the interesection should be like

S1vS1vS1

Er

Hm

No that’s wrong my bad

I don’t know if this approach will work sorry it’s a bit early for me.

[shy moss]

the intersection sould be the base of the cilynder right?

oh thanks

[marsh forge]

Base of the cylinder?

they intersect at all of the wedge points

At a minimum

Because the maps preserve basepoint this will look like one copy of S1

The issue is that depending on how crazy f is

I think the intersection of the two@cylinders can be very different

for example, let f be the identity. Then we get two tori glued along a circle, basically

so the two cylinders only intersect in S^1

However, if I change the map to the swap map, then the two cylinders have a larger intersection because now one cylinder's base is the other cylinders top

so you get S^1vS^1vS^1 if I am visualizing properly

[shy moss]

so if we consider the cilynders union the wedge which is in the base of the cilynder then the intersection is the wedge which is in the base of the cilynder and all the basepoints x I

[marsh forge]

I am struggling to understand what you are describing here. Let's make things as precise as possible. Call the two circles circles a and b. Let $x\in S^1\vee S^1$ be the basepoint of both circles and the wedge. Then let $f$ be as above. If we decompose $T_f$ into two cylinders, we have to think about what the intersection might be. One thing we know for certain is that $f(x)=x$ so that the two cylinders intersect in ${x}\times I/\sim=S^1$. But they might have more intersection. If $f=id$ then this is the only intersection as I said above, but if $f$ is the "swap" map sending a to b and b to a, then the intersection will contain ${x}\times I$ as well as both circles a and b, because $\sim$ glues a to b and vice versa.

[gentle osprey]

SupremumJ

[marsh forge]

Does this make sense @shy moss

I actually do think I know how to solve this formally with van kampen now though (before I just knew from intuition)

[shy moss]

yes

[marsh forge]

Do you want a hint? I think you are on the right path

[shy moss]

yes

[marsh forge]

Hint one: you want to use SvK, but you will have to do casework

Hint two (better hint): ||there are only three possible cases up to homotopy, given by having both a and b in the intersection, neither, or just one of the two without loss of generality||

[shy moss]

Thank you so much

[marsh forge]

Haha I apologize too, when I read your first comment on the problem I noticed the casework and thought "there is no way thats the right method" and then realized it was lol

[charred root]

I’m trying to prove the double cross product formula using exterior algebra, but I’m stuck on $a\times [b\times c] = \star (a\wedge\star(b\wedge c))$

[gentle osprey]

BIGfoot496

[charred root]

How to distribute the Hodge star over wedge product?

I vaguely remember it should be something like $\star(a\wedge b)=\frac{1}{2}\left(P_b(\star a)-P_a(\star b)\right)$

[gentle osprey]

BIGfoot496

[charred root]

Where $P_a$ is some sort of projection parallel to a

[gentle osprey]

BIGfoot496

[charred root]

But whether is it right and how to get a formula for it if it is, I have no clue

[orchid forge]

To be clear, this is only possible if the induced map f* is injective right? If f* = 0 the mapping torus is a cone, and if f*(a) = 0 but f*(b) != 0 then it's two circles attached to one

[marsh forge]

Injectivivity isn’t a big issue but I do think the trivial maps add cases indidnt think of

@shy moss

There are more cases

I mean the answer is completely uniform

But to apply SvK as usual you need some minor casework

The idea is like

f(a) is going to be contained either entirely in one circle or in two, then it matters which circle etc

[shy moss]

i see

[orchid forge]

Injectivity of the induced map

[marsh forge]

I think I covered the case of f(a)=f(b)=a for example in my original

As that would be the single circle intersection case

[orchid forge]

Ok, I'm just not sure what these "cylinders" in T_f look like if f doesn't induce the identity map. If f* = 0 then the cylinders S1 x I in X x I are both (the same) circle {x_0} x I

In the quotient

And then there's the other few cases, if f* is identity, kills one generator, or swaps them

But if that doesn't seem to contradict anything that's been said then it's probably fine

[marsh forge]

Yes maybe your cylinders are secretly cones

In the cases where things are 0

This is a very messy solution to an elegant idea hahaha

[orchid forge]

It's not that bad, you can visualize the mapping torus in each case

[marsh forge]

It is messy relative to the intuition

in terms of having a handful of cases

its not so bad

[orchid forge]

Ok nevermind i think we both missed most of the cases

f* can send each generator to an arbitrary word in a,b

[marsh forge]

That is counted in my cases

Set theoretically that is just one or two of the circles

This suffices for van Kampen

Which then realized the correct relations

Realizes

[orchid forge]

I don't really know what your solution is

Are you saying your cylinders have boundary components (a neighborhood of a circle in X) and the image of that neighborhood?

So the intersection is homotopic to the cone over f(a) \cap f(b)

[orchid forge]

Sorry I keep having to go into classes lol, if you meant that by "cylinder" I think it works

If f*(a) = a^n and f*(b) = b^m for example then in SVK the fundamental groups are <a, c> and <b, c'>, and the generator [x0] for the intersection maps to c^n and (c')^m, which gives you the fundamental group for T_f

Amalgamating over the single relation c^n = (c')^m

I guess it is (homotopic to) a cylinder in the quotient, so I'm pretty sure this is what you meant

[worldly cradle]

Hi this is my first time here so idk if I am doing this correctly

I'm a 4th year math major and I'm taking topology and was wondering if you guys had any tips for how to handle the class

Like important definitions to really get down or stuff like that

[marsh forge]

Point set topology?

@worldly cradle

[worldly cradle]

Its an intro topology class, I think its manifolds

Our book is introduction to topological manifolds by John Lee

[marsh forge]

Oh very different

Interesting

I think there are very few definitions not worth getting down ASAP

Pretty much everything in a class like that would be important

Assuming you’re interested in the material

Otherwise I don’t think the class would be all that different in terms of how you take it

[worldly cradle]

Okay that makes sense, its a lot of set theory right now which isn't bad

[marsh forge]

Well

[worldly cradle]

I'm just not used to using open sets to prove various things so the first homework is a bit rough

[marsh forge]

That’s the confusing part

If you class is on manifolds

Then it won’t be too much of that

Esp if you are using Lee

[worldly cradle]

Yeah I think it may just be the start

[marsh forge]

Point set is boring and a lot of definitions in say munkres aren’t that important

[worldly cradle]

preimage has also been giving me some trouble because inverse and preimage use the same notation

[worldly cradle]

I want to be a HS math teacher so im interested in this stuff because its math but other then that not really

[marsh forge]

Oh

I mean I doubt any of this would be relevant to high school mathematics but it’s cool stuff

[worldly cradle]

Yeah for sure!

I'll probably be here a lot more often with various questions haha

[marsh forge]

@tough imp do you have a good intuition behind geometric points

like what makes them more "geometric" or something

[plain raven]

what is a geometric point

Definition 2.1. A geometric point ξ\xi in SS is a morphism from the spectrum Spec(k¯)Spec(\overline{k}) to SS where k¯\overline{k} is an algebraic closure/separable closure of kk.

this?

[tough imp]

So like

I guess the idea might be that Spec K -> X represents the K points in general

These are like the points that would show up as a variety

Then the geometry is more clear when K is like algebraically closed or something

There’s also a lot of properties about permanence of a property under field extension which you can check over just the algebraic closure or separable closure so maybe that’s what’s going on

I know there’s theorems where you suppose that geometric fibres all have a certain property which is kind of like the fibre over these geometric points

And if those are well behaved then you have certain nice stuff but I don’t know enough specifics to give examples

[marsh forge]

maybe dumb question

but are there more points than geometric points

or is it just that assuming the domain is Spec of an algebraicly closed field is nice

Because obviously we can like take a point of an affine R->k and then compose with inclusion into kbar

[tough imp]

https://mathoverflow.net/questions/546/galois-groups-vs-fundamental-groups/24068#24068

[marsh forge]

and all this should be local right?

[tough imp]

Second answer here gives some shit

Idk if you will understand any bit of it tho

There are points than even points over k

Any map from Spec of a field picks out a certain point

[marsh forge]

sure

[tough imp]

There’s a stacks project thing about points of a scheme

[marsh forge]

Are these not determined by like an affine nbhd

[tough imp]

Yeah

[marsh forge]

oh they aren't

okay

interesting

[tough imp]

Well determined in what sense

[marsh forge]

Like

[tough imp]

I think the answe is yes they are determined affine locally

[marsh forge]

this shows that for an affine scheme the points seen by all fields and seen by alg closed fields should be the same right

and a point of a scheme should be determined by like, an affine nbhd

[tough imp]

Yeah sure

[marsh forge]

Oh wait

hm

[tough imp]

But geometric points are over a specific field

I think the k-points are gonna be in bijection with the geometric points by what you said

[marsh forge]

I see so you want to already be in like, the category of schemes over k or something?

[tough imp]

Yeah

[marsh forge]

ah

I see

that makes more sense

[tough imp]

The definition has a fixed base field

[marsh forge]

But then i could just take k alg closed to start right

does it matter

The reason i am asking is that this comes up in the defn of a smooth morphism of schemes (over k)

[tough imp]

I mean when you pass form the category of k schemes to k-bar scheme

Things do change

[marsh forge]

oh

[true robin]

For a concrete example of a non geometric point, consider maps from specC to spec(C[x]) basically these correspond to maps from C[x] to C, so the preimage of the ideal 0 must be some maximal ideal (x-a)

[tough imp]

Like you might be finite type

Then fail to be finite type

Like if you want to go from k to k-bar

[marsh forge]

I see

[tough imp]

You should really be pulling back your scheme along that map

Or err…

Yeah this is right

It’s like taking a k algebra and tensoring with k-bar

That’s the new object you should probably be considering

Also I think the map does go the wrong way

You don’t have a map Spec k -> Spec k-bad it goes the other way

[marsh forge]

yeah

[tough imp]

So you can’t go from schemes over k to schemes over k-bar by just a composition

You’ll need to take a pullback

[marsh forge]

Right okay

[tough imp]

Which makes sense why we call that base change haha

[marsh forge]

maybe i just need to do some exercises related to the stuff i need to learn

[tough imp]

Lol

Geometric points are kinda

[marsh forge]

This is a side point to me, but like

[tough imp]

Not covered in a standard first year AG thing I think

[marsh forge]

in motivic stuff you really want to consider smooth schemes

[tough imp]

motivic

[marsh forge]

and i have been struggling to get intuition for smooth schemes

[tough imp]

Plsssssss

Oh yeah

Smooth schemes are a little fucked

When it’s a proper variety it’s easier (I mean proper not as in the technical term, I mean like “an actual variety”)

Because it’s the same as being regular

But in general smoothness is a lot trickier to define

If u wait till

Fall qtr is done

I should know a lot more about them

Since Alper’s class is covering this sort of stuff

Étale, smooth, flat stuff

[marsh forge]

I wanna take that class smh

[tough imp]

¯_(ツ)_/¯

[marsh forge]

Oh well, I can take my time with this stuff and blackbox this

Im taking AG this year

[tough imp]

You could ask brofib

[marsh forge]

should be fine

[tough imp]

He’d probably know this stuff better

Altho you might get memed on too hard idk

I know he was doing motivic shit

[marsh forge]

meme'd on?

[tough imp]

Like

[marsh forge]

Yeah he wrote it for his reu paper

[tough imp]

You might want level 2

And you might get level 20

Idk

[marsh forge]

thats fine

[tough imp]

Just ask him to try to dumb it down

[marsh forge]

i dont think that will be super necessary idk

[tough imp]

¯_(ツ)_/¯

I at the very least didn’t understand a single word he said about what motivic cohomology is

[abstract pagoda]

motivation for cohomology

[marsh forge]

a lot of this stuff is emulating AT memes

[tough imp]

To define it alone I was already tapped out

Okay

That helps then

[marsh forge]

@tight agate send me ur reu paper

❤️

I guess i can read it once its on peters website anyway

[limpid leaf]

can someone just clarify what this is asking for me. am i supposed to show that T_B = T_usual? whrer Tb is the topology given by open recntalge basis

[marsh forge]

Yes

[limpid leaf]

k thx

[marsh forge]

(it suffices to show that the given stuff is a basis for T_usual)

[limpid leaf]

so i show that the union of basis elements are equal to T_usual?

no because that is what i originally asked

[silver umbra]

im rly confused by this definition

[limpid leaf]

yeah i dont understand what you're saying actually max

[silver umbra]

it says that the image of a boundary chart cannot intersect the boundary of H

but then it says that a boundary point of M

is a point which is sent to the boundary of H under a boundary chart

[marsh forge]

just show that it is a basis for T_usual

using the basis axioms

bases generate a unique topology

so if something is a basis for one topology and another topology, those two topologies are the same

[limpid leaf]

i tihnk i am just confused by the terminology "a basis for T_usual" because when i look at the defn of a basis in munkres it doesn't actually mention the topology, just the underlying set. Is B a basis for a topology T when we have that for any U in T, if x \in U then there is some b \in B st x \in b and b \subset U?

theres like 4 different ways of phrasing this idea i think and i clearly do not have them straight in my head yet

bc Munkres calls this a "topology T generated by B"?

i think this is what you are saying.

[marsh forge]

This is a definition yes

Given a topology T, a basis B for T is as you described

But pls never use lower case b for a set ever again

[limpid leaf]

I thought the basis axioms were cover + in every intersection of basis is another basis containing x

its B \in \mathcal{B} but u know

[marsh forge]

Munkres might give weird defns

That is what normal people would call a sub basis

A sub basis is sufficient to generate a topology

A basis is more useful

[limpid leaf]

what is a subbasis?

[marsh forge]

.

[limpid leaf]

wtf

[marsh forge]

One sec let me make sure I’m being sane

[limpid leaf]

munkres says a subbasis is a colln of subsets of X whos union equals X

subbasis for a topology on X

[marsh forge]

No

That’s a terrible defn

To anyone sane

[limpid leaf]

🙁

[marsh forge]

This is basis

[marsh forge]

This is sub basis

[marsh forge]

Whatever you described

Is useless

Wikipedia has the good definitions as far as I can tell

[limpid leaf]

wikipedia says the same thing as munkres

• The elements of B cover X, i.e., every element of X belongs to some element in B.
• Given elements B1, B2 of B, for every x in B1 ∩ B2 there is an element B3 in B containing x and such that B3 is a subset of B1 ∩ B2.

[marsh forge]

One sec in a windy place maybe I’m misreading you

Okay

[limpid leaf]

[marsh forge]

My book was fluttering

Very annoying

[marsh forge]

Yes okay if this is what munkres says it is fine

The collection B here

Generates a topology T uniquely

In particular if I start with a topology T

And take a collection of open sets in T

That satisfy

[marsh forge]

This

Then it will also satisfy

[marsh forge]

This

[limpid leaf]

okay and that collection of open sets is called a basis for T

[marsh forge]

Yes

Defining a basis not relative to a topology first

Is insane to me

I forgot munkres did that

So to solve your exercise

It suffices to show that the basis described satisfies

[marsh forge]

This

With respect to the usual open sets

[limpid leaf]

okay so if i am given this open rational square basis (say B) i want to just show that for any U in T_usual, if x \in U then there is some S \in B such that x \in S and S \subset U

[marsh forge]

Yes

[limpid leaf]

okay and if i theoretically did not know this was a basis I would use the basis axioms (cover and intersection) to show that it was before I showed that it generates T_usual

[marsh forge]

If this is true and you know B subset T then you already know it’s a basis (exercise) in the Wikipedia sense

[limpid leaf]

right this is a theorem

[marsh forge]

Ofc if B is not a subset of T this makes no sense to begin with

[limpid leaf]

yea

okay

i tihnk i will work with this more and then i will understand better

[marsh forge]

Okay one last thing

[limpid leaf]

this has straightened things out

[marsh forge]

To maybe help

If you are wanting to build a topology by asserting a basis and taking the topology forced upon you, then the Wikipedia definition is good.

If you already have a topology and want to find a basis, the definition I gave is better

If you have a topology T and a collection of open sets B that satisfy the wiki axioms

Then B will generate T for you

Showing that these are all sensible definitions

[limpid leaf]

right

[marsh forge]

I would almost call the definition relative to a given topology the like, real definition, and a basis that doesn’t have a topology yet some sort of formal basis almost

It’s similar to taking a vector space and finding a spanning basis versus taking a set and freely generating a vector space

[limpid leaf]

right

okay yeah

i sort of get it

thank u very much max

[marsh forge]

You will eventually become so intuitive about it that you will stumble over yourself explaining it to someone else like I just did

[limpid leaf]

lmao

[shy moss]

are the homotopy groups of a kan complex X equal to the homotopy groups of the realization of X?

[marsh forge]

Yes

https://ncatlab.org/nlab/show/simplicial+homotopy+group

see section 3

[shy moss]

thanks

[shy moss]

is every topological space the realization of a simplicial set?

or every cw complex?

[marsh forge]

Up to weak equivalence or homotopy depending on whether you say space or CW

It isn’t true up to homeomorphism

This is the so called homotopy hypothesis

[shy moss]

i have remember that the realization is a cw complex

[marsh forge]

Even when restricted to CW the statement is false if you want homeomorphisms

[orchid forge]

supremum what kind of stuff are you perpetually reading

or perma-studying

[marsh forge]

Oh this role prevents me from seeing the discussion channels

it is exteriorly imposed self control

from the distractions and terrible content therein

[orchid forge]

oh I see, I get the principle

are you reading / researching in topology though?

[marsh forge]

I am a PhD student in topology, yes

[orchid forge]

how specific are you comfortable getting? i am as well, just curious

[marsh forge]

I study at UCSD with Zhouli Xu

this is all pretty public

[orchid forge]

that is pretty specific

[marsh forge]

(I go by MaxJ on the server this name was a meme)

[orchid forge]

well cool, if your research is along the lines of his then I don't know anything about it

[marsh forge]

It is similar

[orchid forge]

(that was always the most likely scenario since I don't know much about anything)

[marsh forge]

I am interested at the moment mostly in like

computational obstructions to non-computational results if that makes sense

like

using down-to-earth if messy computations to prove that certain nice results are impossible

and things like that

[orchid forge]

that sounds good, i'm kind of trying to do something like that right now

obviously the "nice results" are completely different, but poking around for counterexamples with some software and basic reasoning

[marsh forge]

Cool, can I ask what kinds of things you study

[orchid forge]

I'm currently reading about link homology

[marsh forge]

Oh cool

There is something called a massey product which is related to what I study very heavily

[orchid forge]

If it turns out a half decent paper I might commit to it as a research area but for now it's preliminary

[marsh forge]

that first found its home there

idk if youre familiar w it

[orchid forge]

nope

[marsh forge]

the idea is that its like a triple-product that can detect things the cup product cant

so like, the cup product can see linked circles

but the borromean rings are a triplet of rings all linked as a triple, but no two rings are linked

so cup product can't see that

but the massey product can

(in particular the massey product can only be defined when the cup product fails)

[orchid forge]

I see, this is cool

[long hornet]

Hi, I need help seeing why X = (two circles intersecting at two points) is homotopy equivalent to S^1 v S^1 v S^1.

Just drawing

Hopefully the formula wouldn't be too complicated

[marsh forge]

thats a full fledged proof as far as im concerned

[long hornet]

Makes sense

I think I don't have an intuitive grasp yet of what "deformations" I can do

[marsh forge]

You can basically always choose a contractible subspace

and shrink it to a point

most of the transformations are gonna be of this form

you can also just move stuff around and stretch it however

[long hornet]

I see!

[marsh forge]

be careful though

you have to do this one at a time

for example i could choose both edges of the inner venn diagram thing

and try to shrink both

and get the wrong answer

[long hornet]

But it seems hard to me to figure how do the points near the subspace change

[marsh forge]

Well everything else sort of just follows the thing you shrink

but you shouldn't collapse or identify or glue anything extra

[long hornet]

Hmmm..

Okay, in this shape, what if we shrink to the upper intersection point?

[marsh forge]

Do you have an example of something confusing

upper intersection?

do you mean like the x at the top intersection point

[long hornet]

@marsh forge The upper blue point in the left space

[marsh forge]

i see

This would change nothing really

you'd have the point near the intersection go into the intersection

but everything else would stretch with them

and keep the same shape

[long hornet]

Umm

So it would seem like the new inner circle has its vertex on the upper right area

By "vertex" I mean its intersection with the outer circle

[long hornet]

Oh, I think I'm seeing it now!

[marsh forge]

Haha okay I am glad bc i was lost

[long hornet]

Just to be sure, I tried to draw it as a graph

Is my reasoning correct? Here the green lines represent "auxiliary vertices"

The red lines represent the "new" edges that will result from the identification

[marsh forge]

oh my GOD

that diagram man

[long hornet]

ikr!

[marsh forge]

sorry I am not sure i get

what you are contracting

can you highlight it on the original image or smth

[marsh forge]

uh

hm

what subspace though

maybe I can help by redrawing slim's example

[long hornet]

So we have an edge from each green vertex to its neighboring blue point

It will be replaced by the red edge

[marsh forge]

okay so in this example

I start with the space

i highlight a contractible subspace

and then i turn that subspace into a point

can you make a similar diagram for me

(in particular you don't need any arrows)

[long hornet]

Yeah, but I thought thinking of it as a graph will make the argument more convincing.

[long hornet]

Just a sec

Hopefully this is a bit better

[marsh forge]

Okay so the points on the dotted lines like

should not be involved at all

[hollow harbor]

[long hornet]

So slim's idea is to identify v2 and u and v1

[marsh forge]

well

slim shrinks an entire subspace

not just 3 points

you need to shrink the entire line

[long hornet]

I know, and they aren't there. It's just that we had an edge [w2, v2] and then it becomes [w2, u], after the identification is carried out.

[long hornet]

I thought of it as a map defined on the vertices of a simplicial complex and then extended. The result will be the same, right?

[marsh forge]

I would not think about this simplicially honestly

But you diagram looks okay

now that I understand it

[long hornet]

I think I shouldn't, but I just love how we can describe all this "finitely"

[marsh forge]

So what was the other thing

that you tried

[long hornet]

There was one which I found too hard to try: Wedge sum of two circles is htpy eq. to the punctured 2-torus

[marsh forge]

that one you won't be able to do simplicially hahaha

[long hornet]

Also, removing a point is the same as removing a small, closed disk, right? As in, the results are homeomorphic

[marsh forge]

Yeah

for decent spaces

anyway

[long hornet]

I was thinking about manifolds, yeah

[long hornet]

Ugh

Can't I think of a triangulation with the point in one triangle and then remove that triangle?

[marsh forge]

^ good animation

[long hornet]

It is good!

[marsh forge]

Good

[long hornet]

Yes

[long hornet]

I was kidding at first but now that I think of it, I have no other way of answering this question for, say, the Klein bottle

[marsh forge]

The fundamental polygon suffices here

which I wouldn't call a triangulation

[long hornet]

Yeah

By a triangulation I imagined a polygon too, but with triangles on it

[marsh forge]

the triangles are always unecessary

[long hornet]

In the polygon I'm imagining, we don't tell how we glue the edges

Actually we glue more than the edges

[marsh forge]

You only need to glue edges in the fundamental polygon

wait shit

thats a torus

[long hornet]

Yes, but in my polygon, you only need to identify vertices!

This is a torus

[marsh forge]

wtf is this

no

thats terrible lol

[long hornet]

Munkres' algebraic topology book

[honest narwhal]

Are you doing pi_1 stuff?

[long hornet]

This denotes the abstract simplicial complex whose proper faces are {a, b, c}, {a, d, e} etc

[honest narwhal]

You don't need to triangulate stuff if you're doing pi_1 lol

[marsh forge]

Yeah please just

never do this

[honest narwhal]

Just take the space as is

[marsh forge]

you should

change books

no one does this stuff ever

[long hornet]

He does more. I'm currently in the first chapter, which is about simplicial homology (the book doesn't touch homotopy)

[long hornet]

I noticed why it's ~bad~ not optimal. He gives this exercise where you are supposed to compute the homology of n-fold sums of projective planes, but it's easier to see it as a polygon. To "convert" a fundamental polygon to a "triangulated" thing, I had to add too many vertices to keep track of stuff..

[marsh forge]

doing that cellularly takes like 5min

simplicial stuff is just like

outdated

[long hornet]

Hmmm..

Which book starts with homology and uses homological algebra to simplify things, but not too much hom alg as to be unreadable?

[marsh forge]

Starts with?

Probably no modern one

but Hatcher starts with pi1 and you can more or less skip it

[long hornet]

I didn't know that - why?

[marsh forge]

Most books start with pi_1 for some reason

probably to be able to talk about hurewicz

[long hornet]

Aha

[long hornet]

He later treats singular homology and proves that it's isomorphic to simplicial homology

[marsh forge]

Hatcher does delta (an improvement of simplicial)

then cellular and singular

and proves them all equivalent

[long hornet]

I think I'm not comfortable with how handwavy Hatcher can be, but that seems like a thing I can benefit from..

[marsh forge]

I do not think there are any legitimatley "handwavy" proofs in hatcher

outside of maybe ch0

[hollow harbor]

[long hornet]

That's not his first book

I thought it looked great lol

[long hornet]

I will consider it, thanks

Is a graph determined, up to homotopy equivalence, by its homology groups?

[marsh forge]

yes

[long hornet]

Fair enough

[marsh forge]

note that up to homotopy

every graph is just a wedge of circles

so this is straightforward

(both facts are not too hard to prove)

[long hornet]

I thought so but I had no idea how a proof (even handwavy) would go

[marsh forge]

Do you know much graph theory

Here I'll make it a guided exercise

Exercise 1: Given any graph G, it is possible to find a maximal sub-tree T, where by maximal I mean it touches every vertex

Exercise 2: Every graph is homotopy equivalent to a wedge of circles

Exercise 3: Every graph is, up to homotopy, determined by its first homology group

[marsh forge]

Hint, use this lemma somewhere

[orchid forge]

what about infinite graphs

[marsh forge]

I think you can pick a transfinite maximal tree

probably

should be true

[orchid forge]

even in like the complete graph on countably many vertices?

[marsh forge]

I think that should be fine

(the theorem is true regardless, i am just not confident this proof works)

[orchid forge]

oh yeah nevermind, you could take like a horizontal axis and infinitely many parallel vertical axes (for a labeling of the vertices by NxN, but just the chain (n --> n+1) works for N)

[long hornet]

I can vaguely see how every vertex not in T introduces as many "twists" as it has edges to T, or something..?

Like if v is joined to w1,..., w_r in T, then we contract [v, w1] to w1

[marsh forge]

Exercise 1 is just graph theory

no topology

[long hornet]

Wait, maybe I got the definition of a tree wrong

[marsh forge]

no cycles

[long hornet]

Yeah, I was actually doing 2

[marsh forge]

Maybe

exercise 1.5

Every tree is contractible

[long hornet]

A vertex can be connected with many vertices on the tree, right?

[marsh forge]

Yes

[long hornet]

So the idea is to shrink it to a point and see what happens

[marsh forge]

yes

[long hornet]

By focusing my attention on the red vertex, it seems that it becomes homotopic to something like this..?

Ugh, that's not very helpful

@marsh forge Any hints?

[marsh forge]

uh

whats the before?

[orchid forge]

are you trying to do an example of "contracting a maximal spanning tree" for some graph?

[coral pawn]

Why does S \cap U equal that?

[coral pawn]

Nevermind I didn't know centered at p means that the image of p is 0

[dull goblet]

what does basis for open sets mean?

is it like, any open set is contained in the union of the basis sets

or is there another condition too

[little hemlock]

any open set can be written as a union of basis elements, yes

[dull goblet]

finite union?

does it matter

[little hemlock]

just arbitrary.

[dull goblet]

wikipedia has another condition

[empty grove]

Those conditions amount to saying "the set of all unions of these elements is a topology"

[dull goblet]

ok

so do i need to also prove that second condition?

[empty grove]

if you want to convince yourself yes

[dull goblet]

so bases are also supposed to be topologies?

[empty grove]

basis itself is not a topology

it generates a topology by unions

[little hemlock]

wdym by "also?"

[dull goblet]

like i wanna show open sets are in the union of my base sets

is that second condition part of the definition

[little hemlock]

the first condition does not imply on its own that open sets are unions of basis elements

[dull goblet]

oh

so we need equality

(open set) = Union of some base elements

[empty grove]

yes

[dull goblet]

ok

thanks all

when we have a covering of a topology set

would we also have equality?

since its just unioning stuff already in the topology set

in zariski topology atleast

if X = all primes

X < unioning of some primes

equality?

[little hemlock]

you mean a covering of a topological space X? Well if its a covering consisting of subsets of X, then yea, you would have equality. But sometimes we might cover a subspace of a topological space with open sets coming from the larger space

[dull goblet]

ah i see

also this ^

[little hemlock]

[little hemlock]

example of what I was talking about: let X = Spec R and Y = maxSpec R (subspace of maximal ideals of R). the sets X_f = X - V(f) are a basis of Spec R, so they cover X, and therefore cover Y, but since we need not have X = Y, {X_f} is a "cover" of Y that is larger than Y

[dull goblet]

right

makes sense

[little hemlock]

but the sets X_f are also not technically open sets of Y, but the Y \cap X_f are

[dull goblet]

yea

[merry turret]

Hey guys so if I ask for any help with geometry do I ask in this channel?

[gritty widget]

if it fits in

Higher geometry: topology, metric geometry, differential geometry, algebraic geometry, homotopy theory, anime, etc.

then it belongs in here

[merry turret]

Alrighty

[shy moss]

i want to find a presentation of the fundamental group of a simplicial complex X in terms of a spanning tree T of the 1-skeleton of X. First i need to chose a basepoint, say x_0. I think i can descompose X in subspaces A_{u,v}, where u,v are vertices in X and A_{u,v} is the space which consist in the unique line which joins u and v in T, the unique line which joins v and x_0 in T and all the n-simplexes such that one of his vertex is v. Then, is every A_{u,v} simply connected?

or what is the correct way to descompose X and then apply van kampen?

[woven aspen]

does anyone of you have some tips on how to get into clifford / geometric algebra?

[tawdry widget]

Given a connected graph and a base point,why the rank of its fundamental group equals the first Betti number of the graph? And why this first Betti number equals the number of edges minus the number of vertices plus the number of connected components (which is 1 in this case)?

[pearl holly]

well for the first part, I assume that the fundamental group is free. If you abelianize then you get the free abelian group and this is the first homology group. The rank of it is the first Betti number and I think that the rank is preserved under this abelianization in this case?

[empty grove]

Hatcher talks about this toki

[empty grove]

Also, multiposting

[pearl holly]

talks about what?

[empty grove]

Graphs

There's a nice thing they are h eq to

Wrong reply lol

[pearl holly]

yeah in the additional sections right?

or did I miss a page

[empty grove]

Oh maybe

I think this was ch1

You take a maximal subtree

Then you fill in the rest of the argument

[pearl holly]

ah yeah now I remember I think